1.575 g of oxalic acid (COOH)_(2).xH_(2)O are dissolved in water and the volume made up to 250 mL. On titration 16.68 mL of this solution requires 25 mL of N/15 NaOH solution for complete neutralisation. Calculate the value of x.

1.575 g of oxalic acid (COOH)_(2).xH_(2)O are dissolved in water and t

1.	1.575 g of oxalic acid (COOH)_(2).xH_(2)O are dissolved in water and the volume made up to 250 mL. On titration 16.68 mL of this solution requires 25 mL of N/15 NaOH solution for complete neutralisation. Calculate the value of x.
Answer» <html><body><p></p>Solution :Molecular mass of oxalic <a href="https://interviewquestions.tuteehub.com/tag/acid-847491" style="font-weight:bold;" target="_blank" title="Click to know more about ACID">ACID</a>, `(COOH)_2 xH_2O = <a href="https://interviewquestions.tuteehub.com/tag/24-295400" style="font-weight:bold;" target="_blank" title="Click to know more about 24">24</a> + 62 + 2 + 18x =90 + 18x` <br/> Since oxalic acid is a dibasic acid, Eq. mass of oxalic acid `=("Molecular mass")/2` <br/> `=(90 + 8x)/2 = 45 + 9x` <br/> Normality of oxalic acid solution can be calculated from the following relation : <br/> `w=(NEV)/1000` <br/> <a href="https://interviewquestions.tuteehub.com/tag/substituting-1231652" style="font-weight:bold;" target="_blank" title="Click to know more about SUBSTITUTING">SUBSTITUTING</a> the values, we get <br/> `1.575 =(N xx (45 + 9x) xx 250)/1000`<br/> or `N =(1.575 xx 1000)/((45 + 9x) xx 250) = (6.3)/(45 + 9x)` <br/> According to the normality relation, <br/> `underset("oxalic acid")(N_(1)V_(1)) = underset("NaOH")(N_(2)V_(2))` <br/> `6.3/(45 + 9x) xx 16.68 = 1/15 xx 25` <br/> `45 + 9x = (6.3 xx 16.68 xx 15)/25` <br/> `=63.05 = 63` <br/> `9x = 63 - 45 = 18` <br/> or `x=18/9 =2`</body></html>